Object's B height is equal 0 after about 5.5 seconds of flying. To find out if Object A was in the air longer, we can graph the equation for Object's A height and compare it with Object B.
HeightofObjectAh=80t−16t2
The height of the Object A is described with a quadratic equation. Therefore, its graph will be a parabola. We can graph a parabola by finding its vertex and the time-axis intercepts — points in time at which the height of Object A is equal to 0. Finding the intercepts corresponds with solving the following equation.
We found two time-axis intercepts, t=0 seconds and t=5 seconds. Next, we want to find the vertex. Since the graph of the Object's A height equation is a parabola, the vertex's time coordinate will be the midpoint between the time-axis intercepts.
The midpoint between the time-axis intercepts t=0 and t=5 is the point in time t=2.5. Therefore, the time coordinate of the vertex is also t=2.5. To find the height of our vertex, we can substitute its time coordinate into the equation for the height of Object A.
We found that the vertex is the point (2.5,100). Finally, we are ready to graph the height equation of Object A. Let's mark the time-axis intercepts and the vertex on the coordinate plane.
Now, let's sketch the parabola that passes through these points. Since negative time does not make sense to us, we will sketch the graph only for non-negative times. Similarly, we will sketch the graph only for non-negative heights.
Finally, we can graph equations for the heights of both objects and compare the time in the air.
Object's A height is equal to 0 after exactly 5 seconds of flying. Object B was flying for a longer period of time, as it was in the air for about 5.5 seconds.
b Let's look at the graphs of height equations that we made in Part A. This time, let's focus on finding the highest points on these graphs.
We can see that the highest point of the height equation of Object B is at about 70 feet. Object A traveled higher — in Part A we found that the highest point that Object A traveled, the vertex, was at exactly 100 feet.
c To find the times at which Object A was higher than 64 feet, we will first find the times at which the height of Object A is equal to 64 feet.
We can find the exact points by substituting h=64 into Object's A height equation.
h=80t−16t2⇔64=80t−16t2
Let's rewrite the obtained quadratic equation in standard form.
Next, we can calculate the first root using the negative sign and the second root using the positive sign.
t=25±3
t=25−3
t=25+3
t=22
t=28
t=1
t=4
Object A was at 64 feet at t=1 second and at t=4 seconds.
In between t=1 second and t=4 seconds its height was greater than 64 feet!
Object A is was higher than 64 feet in between 1 and 4 seconds.
d Negative times do not sense here, so we can consider only times greater than or equal to 0 seconds. Also, we do not need to analyze negative height.
For times greater than 5 seconds the equation for Object's A height produces negative values. Therefore, we will only consider times less than or equal to 5 seconds. Let's combine these observations and create a domain for the function that describes Object A's height.
Mathleaks uses cookies for an enhanced user experience. By using our website, you agree to the usage of cookies as described in our policy for cookies.
